I can't really do subscripts in this e-mail, so for example Ei(i+1)means E with subscript i,i+1.

Make a new graph by putting the k components in a row -- call them C1,C2, ... , Ck -- with C1 on the left, C2 second from the left, etc,and add one new edge E12 from C1 to C2, one E23 from C2 to C3, ... ,(k-1) in all. Draw each graph without unnecessary edge crossings.Choose the endpoints of Ei(i+1) so that the endpoint in Ci is on theright side of Ci and the one in C(i+1) is on the left. The new graphis connected, has n vertices, e+k-1 edges, and r regions. (If youdraw the edges as I suggested, you don't add any regions. Draw asketch to see this.) Then n - (e+k-1) + r = 2,giving the result.

WDW

On Tue, Oct 23, 2012 at 5:17 PM, Mike Katherein <ytbsearch@yahoo.com> wrote:> G is a planar graph that has n vertices, e edges, r regions, and> k connected components.>> Show that the Euler's Formula for G can be written as: n - e + r = k + 1.> (Remember if G is connected, then k = 1, which means Euler's Formula will be degenerated to n - e + r = 2.)>> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~>> I am a super-newbie to discrete math, and I don't even know where I should start with this question>> I only know that the Eular's theorem for any connected planar graph is n - e + r = 2.>> But I don't know how could I transform that formula to n - e + r = k + 1.>> any hints or helps will be appreciated, thanks for reading this post